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An inductor of 200 mH, capacitor of 400 ...

An inductor of 200 mH, capacitor of `400 mu F` and a resistance of 10 ohm are connected in series to an a.c. source of 50 V of varialbe frequency. Calculate (i) angular frequency at which maximum power dissipation occurs in the circuit and the corresponding value of effective current, and (ii) value of Q factor is the circuit.

Text Solution

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Here, `L = 200 mH = (2)/(10) H`
`C = 400 nu F = 400 xx 10^(-6) F = 4 xx 10^(-4) F`
`R = 10 ohm, E_(v) = 50 V`
(i) Maximum power dissipation occurs in the circuit at resonance, i.e., when
`omega L = (1)/(omega C)`
or `omega = (1)/(sqrt(LC)) = (1)/(sqrt((2)/(10) xx 4 xx 10^(-4)))`
`omega = (1)/(sqrt(80 xx 10^(-6))) = (10^(3))/(8.944)`
`omega = 111.8 rad//s`
`I_(v) = (E_(v))/(Z) = (E_(v))/(R ) = (50)/(10) = 5 A`
(ii) `Q = (1)/(R ) sqrt((L)/(C ))`
`Q = (1)/(10) sqrt((2//10)/(4 xx 10^(-4))) = (1)/(10) xx (100)/(1.47) = 2.237`
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